Question: $ E = \left[\begin{array}{rr}4 & 5 \\ 0 & -1\end{array}\right]$ $ B = \left[\begin{array}{rrr}3 & -2 & 2 \\ 3 & 3 & 0\end{array}\right]$ What is $ E B$ ?
Answer: Because $ E$ has dimensions $(2\times2)$ and $ B$ has dimensions $(2\times3)$ , the answer matrix will have dimensions $(2\times3)$ $ E B = \left[\begin{array}{rr}{4} & {5} \\ {0} & {-1}\end{array}\right] \left[\begin{array}{rrr}{3} & \color{#DF0030}{-2} & \color{#9D38BD}{2} \\ {3} & \color{#DF0030}{3} & \color{#9D38BD}{0}\end{array}\right] = \left[\begin{array}{rrr}? & ? & ? \\ ? & ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ B$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ B$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ B$ , and so on. Add the products together. $ \left[\begin{array}{rrr}{4}\cdot{3}+{5}\cdot{3} & ? & ? \\ ? & ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ B$ and add the products together. $ \left[\begin{array}{rrr}{4}\cdot{3}+{5}\cdot{3} & ? & ? \\ {0}\cdot{3}+{-1}\cdot{3} & ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ B$ and add the products together. $ \left[\begin{array}{rrr}{4}\cdot{3}+{5}\cdot{3} & {4}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{3} & ? \\ {0}\cdot{3}+{-1}\cdot{3} & ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rrr}{4}\cdot{3}+{5}\cdot{3} & {4}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{3} & {4}\cdot\color{#9D38BD}{2}+{5}\cdot\color{#9D38BD}{0} \\ {0}\cdot{3}+{-1}\cdot{3} & {0}\cdot\color{#DF0030}{-2}+{-1}\cdot\color{#DF0030}{3} & {0}\cdot\color{#9D38BD}{2}+{-1}\cdot\color{#9D38BD}{0}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rrr}27 & 7 & 8 \\ -3 & -3 & 0\end{array}\right] $